In general, the use of LEDs for illumination purposes is known. A problem with LEDs is the power supply; it is noted that the power supply in a car is provided by the car's battery, typically providing a voltage in the order of 6 V or 12 V or 24 V. For a LED to produce light, it requires a current to pass through it in one direction (from anode to cathode); current flow in the opposite direction is blocked. When driven with current having the correct direction, a voltage drop develops over the LED which is substantially independent of the LED current. Within margins, the LED current can be varied, and the light output will be substantially proportional to this current. When it is desirable to produce more light than one LED can generate, it is possible to combine multiple LEDs. The LEDs can be arranged in a series arrangement, which would require a higher voltage drop at the same current, or the LEDs can be arranged in a parallel arrangement, which requires more current at the same voltage drop. Thus, the costs of power supply increase. Combinations of series arrangement and parallel arrangement are also possible.
A relatively simple and cheap way of powering a plurality of LEDs is to connect all LEDs in series and to connect this string to the battery, having a current limiting resistor in series. A problem when powering a LED or a string of LEDs directly from a car battery is that the supply voltage may change substantially with time. FIG. 1 is a graph showing a relationship between supply voltage and LED current. A horizontal dotted line 11 represents the required voltage drop, also indicated as forward voltage, over a string of LEDs. Curve 12 represents battery voltage. Assume that the horizontal axis represents time. Assume that in period A the car's motor is off and the battery voltage is nominal and higher than the required voltage drop: the LEDs pass a current (curve 13) and light is generated. The difference between supply voltage and voltage drop is accommodated by the series resistor, and involves loss of energy by dissipation in the resistor. Assume that in period B the car's motor is being started so that the battery voltage drops and becomes lower than the required voltage drop: the LEDs can not pass current and can not generate light. Assume that in period C the motor is running and the battery voltage is higher than nominal: the series resistor needs to accommodate more voltage, thus the power dissipated in the resistor will increase.